# Math Help - Linear Algebra: Eigenvalues question

1. ## Linear Algebra: Eigenvalues question

Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!

2. Originally Posted by openustudent
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!
It will help if you tell us what your notation means. what does p(A) = 1 mean?

3. p(A) is the rank of the matrix A (in our case - 1) and tr(A) is the trace of the matrix A.

4. Originally Posted by openustudent
Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

Thanks!
Since the rank is 1, there exists vectors $x$ and $y$ such that $A = xy^T$(Prove this!)

This matrix has atmost one non zero eigenvalue. Clearly $x$ is an eigen vector corresponding to the non-zero(if it is...) eigenvalue $x^T y$.

But observe that $\text{Tr} (xy^T) = x^Ty$, thus all eigenvalues of A are 0.

There are two statements though that I did not understand:

This matrix has atmost one non zero eigenvalue
How come ?

Clearly is an eigen vector corresponding to the non-zero(if it is...) eigenvalue .
Why is this?

6. Originally Posted by openustudent
How come ?
Why is this?
Observe that $Ax = (xy^T)x = x(y^Tx) = (y^Tx)x$. If you still cant see it, observe that $y^Tx$ is a scalar.
Let us say it had an eigenvector z, corresponding to a eigen value $\alpha$ ,then $Az = \alpha z \implies (xy^T)z = \alpha z \implies (y^T z) x = \alpha z$ , Thus if $\alpha \neq 0$, z is in the span of x.